Answer:
[tex]\mathbf{\Delta V = -0.63 \%}[/tex]
Contraction
Explanation:
From the given information;
Above 88° C
zirconium has a BCC crystal structure with a = 0.332 nm
Below this temperature
zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm
the volume of BCC can now be:
[tex]V_{BCC} = (a)^3[/tex]
[tex]V_{BCC} = (0.332 \ nm)^3[/tex]
[tex]V_{BCC} =0.03660 \ nm^3[/tex]
the volume of HCP can now be:
[tex]V_{HCP} = (a)^2 (c) cos \ 30^0[/tex]
[tex]V_{HCP} = (0.2978)^2 (0.4735) \ cos 30[/tex]
[tex]V_{HCP} =0.03637 \ nm^3[/tex]
Ths; the volume percent change when BCC zirconium transforms to HCP zirconium can be calculated as:
[tex]\Delta V = \dfrac{V_{HCP}-V_{BCC}}{V_{BCC}} * 100 \%[/tex]
[tex]\Delta V = \dfrac{0.03637 \ nm^3-0.03660\ nm^3}{0.03660\ nm^3}} * 100 \%[/tex]
[tex]\mathbf{\Delta V = -0.63 \%}[/tex]
Hence; it is contraction due to what the negative sign portray, The negative sign signifies that there is contraction during cooling
A 25.00 mL solution of sulfuric acid (H2SO4) is titrated to phenolphthalein end point with 27.00 mL of 1.700 M KOH. Calculate the molarity of the acid solution? H2SO4(aq) + 2KOH(aq) o K2SO4(aq) + 2H2O(l)
Answer:
0.9180 M
Explanation:
Step 1: Write the balanced equation
H₂SO₄(aq) + 2 KOH(aq) ⇒ K₂SO₄(aq) + 2 H₂O(l)
Step 2: Calculate the reacting moles of KOH
27.00 mL of 1.700 M KOH react. The reacting moles of KOH are:
[tex]0.02700L \times \frac{1.700mol}{L} = 0.04590mol[/tex]
Step 3: Calculate the reacting moles of H₂SO₄
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 0.04590 mol = 0.02295 mol.
Step 4: Calculate the molarity of H₂SO₄
0.02295 moles of H₂SO₄ are in 25.00 mL of solution. The molarity of the acid solution is:
[tex]M = \frac{0.02295 mol}{0.02500} = 0.9180 M[/tex]
Refer to the example about diatomic gases A and B in the text to do problems 20-28.
It was determined that 1 mole of B2 is needed to react with 3 moles of A2.
How many grams in one mole of B2?
__g
Answer:
28g.
Explanation:
Hello,
In this case, considering the statement, we can infer that the monoatomic atomic mass of B is 14 g in one one mole. In such a way, since it is diatomic, we can notice that one mole of B2, is having 28 g of B2, as monoatomic atomic mass is considered twice.
Regards.
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution
Answer:
Concentration of the H₂SO₄ solution is 0.0737 M
Explanation:
Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:
H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O
From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.
mole ratio of acid to base, nA/nB = 1:2
Concentration of the base, Cb = 0.1311 M
Volume of base, Vb, = 28.11 mL
Concentration of acid, Ca = ?
Volume of acid, Va + 25.0 mL
Using the formula, CaVa/CbVb = nA/nB
making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB
substituting the values into the equation
Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M
Therefore, concentration of the H₂SO₄ solution is 0.0737 M
The solvent was propanone. Which of the three basic colours is most soluble in propanone?
Answer:
Red dye
Explanation:
In the given question, the complete question has not been provided but the propanone is used as a solvent in paper chromatography. The paper chromatography was performed for the black ink in which the black ink got separated in the red, blue and yellow colour.
From these three colours that are red, blue and yellow, the dye which is most soluble in propanone was red as red colour moved the most in the given chromatogram and the dye which travelled the most is most soluble in propanone.
Thus, red dye is the correct answer.
A 13.0-L helium tank is pressurized to 26.0 atm. When connected to this tank, a balloon will inflate because the pressure inside the tank is greater than the atmospheric pressure pushing on the outside of the balloon. Assuming the balloon could expand indefinitely and never burst, the pressure would eventually equalize causing the balloon to stop inflating. What would the volume of the balloon be when this happens? Assume atmospheric pressure is 1.00 atm. Also assume ideal behavior and constant temperature. i got 338L for he whole thing but that is the volume of the entire sample of helium. But you need to consider that 13.0 liters of that is still in the 13.0-L tank. A helium tank is able to inflate balloons if the inside pressure is greater than the atmospheric pressure. can you explain how to do this
Answer:
The volume of the ballon is 325L.
Explanation:
Boyle's law express that the pressure of a gas is inversely proportional to its volume. That means if the pressure increases, the volume decreases. The formula is:
P₁V₁ = P₂V₂
Where P represents pressure and V volume of 1, initial state and 2, final state of the gas.
In the problem, the volume of the tank is 13.0L and the final pressure of the ballon is 1atm -The atmospheric pressure-. As 1atm of gas is in the ballon, the pressure of the tank is 26.0atm - 1.0atm = 25.0atm.
Replacing in Boyle's law expression:
25.0atm*13.0L = 1atmV₂
325L = V₂
The volume of the ballon is 325L.
The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.65 mM2.65 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
[tex]K_c = e^{-0.67}[/tex]
[tex]K_c =[/tex] 0.511
Now using the equilibrium constant [tex]K_c[/tex]
[tex]K_c = \dfrac{[F6P]}{[G6P]}[/tex]
[tex]0.511 = \dfrac{[F6P]}{[2.65]}[/tex]
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
Morphine is a well known pain killer but is highly addictive. The lethal dose of morphine varies from person to person based on their body weight and other factors but is somewhere near 70 mg. Calculate the number of millimoles of carbon atoms in 71.891 mg sample of morphine. Report your answer to the third decimal place.
Answer:
0.252 milimoles
Explanation:
To convert mass of a substance to moles it is necessary to use the molar mass of the substance.
The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:
C: 17*12.01g/mol = 204.17g/mol
H: 19*1.01g/mol = 19.19g/mol
N: 1*14g/mol = 14g/mol
O: 3*16g/mol = 48g/mol.
204.17 + 19.19 + 14 + 16 = 285.36g/mol
Thus, moles of 71.891 mg = 0.071891g:
0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles
As 1 mole = 1000 milimoles:
2.5193x10⁻⁴ moles = 0.252 milimoles
Prepare a solution that is 0.1 M acetic acid and 0.1 M sodium acetate by measuring out 5.0 mL of the 1.0 M acetic acid solution and 5.0 mL of the 1.0 M sodium acetate solution in a 100 mL graduated cylinder, diluting the 10.0 mL to a final volume of 50.0 mL with deionized water, and then stirring. Pour this solution into a clean, dry 100 mL breaker. By knowing that the Ka for acetic acid is 1.8 x -5 10 , calculate the theoretical pH of the solution.
Answer:
4.74
Explanation:
It is possible to find pH of a buffer (The mixture of a weak acid: Acetic acid, with its conjugate base: Sodium acetate) using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka of the weak acid, [A⁻] concentration of the conjugate base and [HA] concentration of the weak acid
pKa of acetic acid is -log 1.8x10⁻⁵ = 4.74
The concentration of both, acetic acid and sodium acetate is 0.1M. Replacing in H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
pH = 4.74 + log₁₀ [0.1] / [0.1]
pH = 4.74 + log₁₀ 1
pH = 4.74
Theoretical pH is 4.74
Thermal decomposition of 5.00 metric tons of limestone to lime and carbon dioxide requires 9.00 × 106 kJ of heat. Convert this energy to (a) joules; (b) calories; (c) British thermal units. Give your answers in scientific notation.
Answer:
Take a look at the attachment below
Explanation:
Hope that helps!
chemical equation for potassium sulfate and lead(II) acetate
Answer:
K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂
A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:
[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]
A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.
In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.
Therefore, the chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:
[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]
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Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hydrogen carbonate, NaHCO3, are combined. Include physical states.
Answer:
[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]
Explanation:
In this case, we will have a neutralization reaction. We have a base ([tex]HI[/tex]) and a base ([tex]NaHCO_3[/tex]). Additionally, we have a strong acid and a strong base, therefore both will be soluble on water, so we will have an aqueous state for these compounds. If we will have a neutralization reaction, we will have as a salt as a product. With this in mind the reaction would be:
[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]
All the sodium salts are soluble in water, therefore we will have an aqueous state. Water is a liquid and carbon dioxide is a gas.
I hope it helps!
Calculate the change in enthalpy associated with the combustion of 322 g of ethanol. C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH∘c=−1366.8kJ/mol
Answer: The change in enthalpy associated with the combustion of 322 g of ethanol is [tex]-9567.6kJ[/tex]
Explanation:
To calculate the number of moles we use the equation:
[tex]\text{Moles}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of ethanol}=\frac{322g}{46g/mol}=7moles[/tex]
The balanced chemical reaction is:
[tex]C_2H_5OH(l)+3O_2\rightarrow 2CO_2(g)+3H_2O(l)[/tex] [tex]\Delta H=-1366.8kJ/mol[/tex]
Given :
Energy released when 1 mole of ethanol is combusted = 1366.8 J
Thus Energy released when 7 moles of ethanol is combusted =[tex]\frac{1366.8}{1}\times 7=9567.6kJ[/tex]
Thus the change in enthalpy associated with the combustion of 322 g of ethanol is [tex]-9567.6kJ[/tex]
The change in enthalpy associated with the combustion is -9567.6KJ
Calculation of change in enthalpy:Since there is 322g of ethanol
Also, there is the chemical equation i.e.
C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH∘c=−1366.8kJ/mol
So, the change should be
= -1366.8kJ *7/1
= -9567.6KJ
Since Energy released at the time when 1 mole of ethanol is combusted = 1366.8 J
So, here Energy released when 7 moles of ethanol is combusted
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The following reaction: NO2 (g) --> NO (g) 1/2 O2 (g) is second-order in the reactant. The rate constant for this reaction is 3.40 L/mol*min. Determine the time needed for the concentration of NO2 to decrease from 2.00 M to 1.50 M.
Answer:
t = 0.049 mins or 2.94 secs
Explanation:
For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;
Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.
Using the integrated rate law:
I/[NO2]t - 1/[NO2]o = Kt
Where K = 3.40 L/mol/min
[NO2]t = 1.5 mol/L
[N02]o = 2.0 mol/L
t = ?
Making t subject of formula;
t = (1/[NO2]t - 1/[NO2]o) / K
t = (1/1.5 - 1/2.0)/3.40
t = 0.049 mins or 2.94 secs
The de Broglie wavelength of an electron with a velocity of 6.90 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-28 g.
Answer:
Explanation:
use this equation and solve for wavelength (λ)
λ = h/mv.
where h is plancks constant 6.63 × 10−34 J·s
m = mass of lectron
v = velcoeity of electron
Of Sr or Ba , the element with the higher first ionization energy is
Answer:
Sr
Explanation:
Sr has an ionization of 550 whereas Ba has an ionization of 503
The most common isotopic forms of hydrogen are ordinary hydrogen (1H) and deuterium (2H), which have percent compositions of 99.98% and 0.0115%, respectively. Convert the percent isotopic composition value of 2H to decimal form.
Answer:
0. 000115
Explanation:
A percentage is defined as a ratio with a basis of 100 as total substance. Convert a percentage to decimal implies to divide the percentage in 100 because decimal form has as basis 1.
For the isotopic forms:
1H: 99.98% → As percent.
99.98% / 100 = 0.9998 → As decimal form.
2H: 0.0115% → As percent.
0.0115% / 100 = 0. 000115→ As decimal form.
The percent should be 0. 000115
The calculation is as follows:For the isotopic forms:
1H: 99.98% → As percent.
Now
[tex]99.98\% \div 100[/tex]= 0.9998 → As decimal form.
Now
2H: 0.0115% → As percent.
And,
[tex]0.0115\% \div 100[/tex]= 0. 000115→ As decimal form.
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The complete ionic equation for the reaction of aqueous sodium hydroxide with aqueous nitric acid is
Answer and Explanation:
Sodium hydroxide (NaOH) is a strong base and nitric acid (HNO₃) is a strong acid. That means that they dissociates in water by giving the ions:
NaOH ⇒ Na⁺(ac) + OH⁻(ac)
HNO₃ ⇒ H⁺(ac) + NO₃⁻(ac)
The reaction between an acid and a base is called neutralization. In this case, HNO₃ loses its proton and it is converted in NO₃⁻ (nitrate anion). NaOH loses its hydroxyl anion (OH⁻) by giving Na⁺ cations.
Na⁺ cations with NO₃⁻ anions form the salt NaNO₃ (sodium nitrate); whereas H⁺ and OH⁻ form water molecules. The complete equation is the following:
HNO₃(ac) + NaOH(ac) ⇒ NaNO₃(ac) + H₂O(l)
The ionic equation is:
H⁺(ac) + NO₃⁻(ac) + Na⁺(ac) + OH⁻(ac) ⇄ Na⁺(ac) + NO₃⁻(ac) + H₂O(ac)
If we cancel the repeated ions at both sides of the equation, it gives the following ionic reaction:
H⁺(ac) + OH⁻(ac) ⇄ H₂O(ac)
Consider the three isomeric alkanes n-hexane,2,3-dimethylbutane, and 2-methylpentane. Which of the following correctly lists these compounds in order of increasing boiling point
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
b. 2-methylpentane
c. 2-methylpentane < 2,3-dimethylbutane
d. n-hexane < 2-methylpentane < 2,3-dimethylbutane
e. n-hexane < 2,3-dimethylbutane < 2-methylpentane
Answer:
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
Explanation:
The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.
The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.
n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.
The volume of a sample of water is 2.5 mL the volume of the sample in liters is
Answer:
0.0025Litters
Explanation:
2.5ml= 2.5x10^-3l
2.5ml= 0.0025l
Answer:
AAAAAAAA
Explanation:
find the mass of h2 produced Binary compounds of alkali metals and hydrogen react with water to produce H2(g). The H2H2 from the reaction of a sample of NaH with an excess of water fills a volume of 0.505 L above the water. The temperature of the gas is 35 ∘C∘C and the total pressure is 755 mmHg
Answer: Mass of hydrogen produced is 0.0376 g.
Explanation:
The reaction equation will be as follows.
[tex]NaH(aq) + H_{2}O(l) \rightarrow H_{2}(g) + NaOH(aq)[/tex]
Now, formula for total pressure will be as follows.
[tex]P_{total} = P_{H_{2}} + P_{H_{2}O}[/tex]
Hence, [tex]P_{H_{2}} = P_{total} - P_{H_{2}O}[/tex]
= 755 mm Hg - 42.23 mm Hg
= 712.77 mm Hg
[tex]P_{H_{2}} = \frac{712.77 \times 1 atm}{760 mm Hg}[/tex]
= 0.937 atm
Now, we will calculate the moles of [tex]H_{2}[/tex] as follows.
[tex]P_{H_{2}}V = nRT[/tex]
[tex]0.937 atm \times 0.505 L = n \times 0.0821 \times 308.15 K[/tex]
n = [tex]\frac{0.473}{25.29}[/tex] mol
= 0.0187 mol
Therefore, mass of [tex]H_{2}[/tex] will be calculated as follows.
[tex]m_{H_{2}} = \frac{0.0187 mol \times 2.0158 g}{1 mol}[/tex]
= 0.0376 g
Thus, we can conclude that mass of hydrogen produced is 0.0376 g.
The quantum theory of energy levels within atoms was aided by:
study of the sun's light spectrum
emission line spectra of various elements
alpha particles
gamma rays
Answer:
it was based on the studies by emission line spectra of various elements
Explanation:
Answer:
The answer is: emission line spectra of various elements
Explanation:
PdPd has an anomalous electron configuration. Write the observed electron configuration of PdPd. Express your answer in complete form in order of orbital filling. For example, 1s22s21s22s2 should be entered as 1s^22s^2. View Available Hint(s)
Answer:
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
Explanation:
Palladium is a chemical element with the symbol Pd and atomic number 46.
The electronic configuration is;
[Kr] 4d¹⁰
The full electronic configuration observed for palladium is given as;
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
The reason for for the anomlaous electron configuration is beacuse;
1. Full d orbitals are more stable than partially filled ones.
2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.
How many types of endoplasmic recticulum are there in a cell?
Answer:
Two
Explanation:
2
rough endoplasmic reticulum
smooth endoplasmic reticulum
A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The specific heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g
Answer:
35.578g or 36g if you round
Explanation:
Q=mc ∆∅ where ∅ is temperature difference
1160= m x 1.716 x (42-23)
m = 1160/ 1.716 x19
m=35.578g
m = 36g to nearest whole number
Answer: C. 36 g
Explanation: I got this right on Edmentum.
An experiment calls for 10.0 mL of bromine (d = 3.12 g/mL). Since an accurate balance is available, it is decided to measure the bromine by mass. How many grams should be measured out? Multiple Choice 3.21 32.1 3.12 31.2 0.312
Answer:
31.2g
Explanation:
The following data were obtained from the question:
Volume of bromine = 10mL
Density of bromine = 3.12 g/mL
Mass of bromine =...?
The Density of the substance is related to it's mass and volume by the following equation:
Density = Mass /volume
With the above equation, we can calculate the mass of bromine as follow:
Density = Mass /volume
Volume of bromine = 10mL
Density of bromine = 3.12 g/mL
Mass of bromine =...?
Density = Mass /volume
3.12 = Mass /10
Cross multiply
Mass of bromine = 3.12 x 10
Mass of bromine = 31.2g
Therefore, the mass of bromine is 31.2g
A solution of nitrous acid and potassium nitrite acts as a buffer due to reactions that occur within the solution when a strong acid or a strong base is added.
(a) Write the net ionic equation for the reaction that occurs in this buffer to react away any added HCl (aq).
(b) Write the net ionic equation for the reaction that occurs in this buffer to react away any added NaOH (aq).
Answer:
a. NO₂⁻ + H⁺ → HNO₂
b. HNO₂ + OH⁻ → NO₂⁻ + H₂O
Explanation:
A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.
The buffer of the problem is HNO₂/NO₂⁻ where nitrous acid is the weak acid and NO₂⁻ is its conjugate base.
a. When a acid is added to a buffer as the buffer of the problem, the conjugate base will react with the acid, to produce the weak acid, thus:
NO₂⁻ + HCl → HNO₂ + Cl⁻
Ionic equation is:
NO₂⁻ + H⁺ + Cl⁻ → HNO₂ + Cl⁻
In the net ionic equation, you avoid the ions that don't react, that is:
NO₂⁻ + H⁺ → HNO₂b. In the same way, the weak acid will react with the strong acid producing water and the conjugate base, thus:
HNO₂ + NaOH → NO₂⁻ + H₂O + Na⁺
The ionic equation is:
HNO₂ + Na⁺ + OH⁻ → NO₂⁻ + H₂O + Na⁺
And the net ionic equation is:
HNO₂ + OH⁻ → NO₂⁻ + H₂OA certain reaction with an activation energy of 155 kJ/mol was run at 495 K and again at 515 K . What is the ratio of f at the higher temperature to f at the lower temperature
Answer:
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Explanation:
Using the sum of Arrhenius equation you can obtain:
ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)
Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)
Replacing:
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
Where 2 represents the state with the higher temperature and 1 the lower temperature.
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
ln (f₂/f₁) = 1.4626
f₂/f₁ = 4.32
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Which statement describes both homogeneous mixtures and heterogeneous mixtures?
Answer:
both are the types of mixture and both are impure substances that donot have fixed composition and the composition of constituents is not uniform
Answer:
Their components van be separated by physical processes
Explanation:
Out of the answers im given, it makes the most sense. I would double check before submitting though
solution to a solution of D gives a white precipitate, F.
a State the names of D, E and F.
D is a green crystalline solid that dissolves in water to give a very pale
green solution. Addition of sodium hydroxide solution to a solution of D
produces a green precipitate, E, which turns orange-brown around the top
after standing in air. Addition of dilute hydrochloric acid and barium chloride
Naturally occurring sulfur consists of four isotopes: 32S (31.97207 u, 95.0%); 33S (32.97146 u, 0.76%); 34S (33.96786 u, 4.22%); and 36S (35.96709 u, 0.014%). Calculate the average atomic mass of sulfur in atomic mass units.
Answer:
32.062
Explanation:
The following data were obtained from the question:
Mass of isotope A (32S) = 31.97207 u
Abundance of isotope A (A%) = 95.0%
Mass of isotope B (33S) = 32.97146 u Abundance of isotope B (B%) = 0.76%
Mass of isotope C (34S) = 33.96786 u
Abundance of isotope C (C%) = 4.22%
Mass of isotope D (36S) = 35.96709 u Abundance of isotope D (D%) = 0.014%
Average atomic mass of S =..?
The average atomic mass of sulphur, S can be obtained as follow:
Average atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100] + [(Mass of C x C%)/100] + [(Mass of D x D%)/100]
Average atomic mass of sulphur =
[(31.97207 x 95)/100] + [(32.97146 x 0.76)/100] + [(33.96786 x 4.22)/100] + [(35.96709 x 0.014)/100]
= 30.373 + 0.251 + 1.433 + 0.005
= 32.062
Therefore, the average atomic mass of sulphur is 32.062